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Answered by
2
Answer:
1
(
1+2Logx
1−2Logx
)+tan
−1
(
1−6Logx
3+2Logx
)
y=tan
−1
1−(
1+2Logx
1−2Logx
)(
1−6Logx
3+2Logx
)
1+2Logx
1−2Logx
+
1−6Logx
3+2Logx
=tan
−1
(
(1+2Logx)(1−6Logx)−(1−2Logx)(3+2Logx)
(1−6Logx)(1−2Logx)+(3+2Logx)(1+2Logx)
)
=tan
−1
1−4Logx−12Log
2
x−(3−4Logx−4Log
2
x)
(1−8Logx+12Log
2
x+3+8Logx+4Log
2
x)
=tan
−1
(
−2−8Log
2
x
4+16Log
2
x
)
y=tan
−1
(−
2
1
)
y=c
dx
dy
=0
dx
2
d
2
y
=0
Answered by
10
Answer:
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